Finding eigenvalues of a linear operator

Question

This is part of a linear algebra assignment, and I was hoping that I could gain some hints.

Given a standard basis of $\mathbb{C}^n$, $\{e_1,e_2, \cdots, e_n\},$ I am given a linear operator $S: \mathbb{C}^n \rightarrow \mathbb{C}^n$ such that $Se_1 = e_2, Se_2, e_3, \ldots, Se_{n} = e_1$. I was able to deduce that it is a unitary operator; hence its eigenvalues must be of magnitude 1 and that the eigenvectors are orthogonal to the standard complex inner product.

But I now want to find an explicit expression for all the eigenvalues. I wrote out the matrix representation of the linear operator w.r.t the standard base which gives $$ S = \begin{bmatrix} 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$ I was wondering if there is a way to write a generic experession for the characteristic polynomial.

Answer 1

Note that $S$ is a companion matrix from which we can read off the characteristic polynomial $\chi_S(\lambda) = \lambda^n-1$ from which the eigenvalues follow.

To see the formula for $\chi_S$, note that $\det (\lambda I -S) = \lambda \det S_{ [2..n, 2..n]} + (-1)^n \det S_{ [2..n,1..n-1]}$ and a quick glance at the matrix shows $\det S_{ [2..n, 2..n]} = \lambda^{n-1}$ and $\det S_{ [2..n,1..n-1]} = (-1)^{n-1}$ from which the formula follows.

Answer 2

Suppose $v=(v_1,...,v_n)^t$ is an eigenvector. Then one can see that $v_{i+1}=\lambda v_i$ for $i<n$ and $v_n=\lambda v_1$. This means the entries are in a geometric progression with ratio $\lambda$ but since it cycles back, $\lambda^n=1$, so the eigenvalues are $n$th roots of unity. One can see that each root of unity can occur as an eigenvalue, and since there are $n$ of them, the characteristic polynomial is thus determined as $x^n-1$.